[Math] Conjugate Mobius Transformations Are Invariant

complex-analysis

This is for homework and I was hoping to get some help with clarifying some of the concepts.

The problem is as follows:
Let $f,g$ be conjugate Mobius transformations, say $g=h^{-1} \circ f \circ h$ for a Mobius transformation $h$. Verify $K$ is a $g-invariant$ circle in $\mathbb{C}\cup\{\infty\}$ iff $h(K)$ is invariant under $f$.

My questions are:

  1. What does it mean for $f,g$ to be conjugate.

  2. What does it mean for K to be a $g-invariant$ circle.

If I can understand these concepts I feel that I can do the rest. Any help would be appreciated.

Best Answer

question 1. If $g=h^{-1}\circ f\circ h$ (or equivalently $f=h\circ g\circ h^{-1}$) for a Mobius transformation $h$, then $f$ and $g$ are conjugate.

question 2. A circle $K$ is $g$-invariant iff the image $g(K)$ is the same circle $K$.

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